Question

Factor: \[\frac{ 2p-7 }{ 9 }^{3}\div \frac{ 7-2p }{ 3 }^{4}\]

Answer 1

Are you sure that isn't
\[\frac{(2p-7)^3}{9} \div\frac{(7-2p)^4}{3}\]?

Answer 2

it is...

Answer 3

Okay, what you wrote is very different...

Answer 4

Do you remember how to divide by a fraction?

Answer 5

you have to flip it because it changes to multiplication and then factor

Answer 6

Flip it and multiply, right.

If you flip and multiply, do you see any factors you can cancel?

Answer 7

729 and 81 have 9 in common

Answer 8

Where are you getting 729 and 81 from?

If we flip and multiply, we have
\[\frac{(2p-7)^3}{9}*\frac{3}{(7-2p)^4}\]What cancellation can you do right off the bat?

Answer 9

9^3 and 3^4

Answer 10

Unless you miswrote the problem in another way, the exponent isn't applied to the 3 and the 9...

Answer 11

You said it was \[\frac{(2p-7)^3}{9} \div \frac{(7-2p)^4}{3}\]Are you now saying it is really
\[(\frac{(2p-7)}{9})^3 \div (\frac{(7-2p)}{3})^4\]

Answer 12

Why did you say the other version was right?

Answer 13

it doesnt matter now..thats the problem...can we just solve it?

Answer 14

Well, no matter, I guess.

After we flip and multiply, we get
\[( \frac{2p-7}{9} )^3 * ( \frac{3}{7-2p} )^4\]What do you see to cancel there?

Answer 15

you cant cancel until you multiply all that stuff....

Answer 16

Sure you can. what about the 9 and the 3 to various powers?

Answer 17

What is \[\frac{3^4}{9^3}\]

Answer 18

big numbers

Answer 19

Isn't that just \[\frac{3*3*3*3}{9*9*9}\]?

Answer 20

yes

Answer 21

And that simplifies to...

Answer 22

\[9^3 = (3^2)^3 = 3^6\]

Answer 23

\[\frac{3^4}{3^6} = 3^{4-6} = 3^{-2} = \frac{1}{3^2} = \frac{1}{9}\]

Answer 24

Or if you're more into the visual, each pair of 3's in the numerator cancels out a 9 in the denominator, leaving \(\dfrac{1}{9}\)

Answer 25

what about the 2p-7 and 7-2p?